Coordinate geometry

Start by calculating the gradient $m$ of the line and then the $y$-intercept $(c)$. Recall that the the equation of a straight line is of the form $y = mx + c$.

To calculate the equation of the straight line, we first calculate the gradient ($m$) of the line $AB$:

Secondly, we calculate the value of the y-intercept ($c$) of the line $AB$. We do this by substituing a point $A$ into the general form for a straight line:

Therefore, the equation of the line $AB$ is as follows:

Start by calculating the gradient $m$ of the line and then the $y$-intercept $(c)$. Recall that the the equation of a straight line is of the form $y = mx + c$.

To calculate the equation of the straight line, we first calculate the gradient ($m$) of the line $AB$:

Secondly, we calculate the value of the y-intercept ($c$) of the line $AB$. We do this by substituing a point $A$ into the general form for a straight line:

Therefore, the equation of the line $AB$ is as follows:

Start by calculating the gradient $m$ of the line and then the $y$-intercept $(c)$. Recall that the the equation of a straight line is of the form $y = mx + c$.

To calculate the equation of the straight line, we first calculate the gradient ($m$) of the line $AB$:

Secondly, we calculate the value of the y-intercept ($c$) of the line $AB$. We do this by substituing a point $A$ into the general form for a straight line:

Therefore, the equation of the line $AB$ is as follows:

The general form of the equation of a straight line:

In the question statement we are given $m$, so the only unknown is $c$.

We need to calculate the value of the $y$-intercept ($c$) of the line. We do this by substituing the coordinates of the given point into the general form for a straight line:

Therefore, the equation of the straight line is:

The general form of the equation of a straight line:

In the question statement we are given $m$, so the only unknown is $c$.

We need to calculate the value of the $y$-intercept ($c$) of the line. We do this by substituing the coordinates of the given point into the general form for a straight line:

Therefore, the equation of the straight line is:

The general form of the equation of a straight line:

In the question statement we are given $m$, so the only unknown is $c$.

We need to calculate the value of the $y$-intercept ($c$) of the line. We do this by substituing the coordinates of the given point into the general form for a straight line:

Therefore, the equation of the straight line is:

To calculate the equation of the straight line, we first calculate the gradient $m$ of the line:

Secondly, we calculate the value of the $y$-intercept ($c$). We do this by substituing the coordinates of point $A$ into the general form for a straight line:

Therefore, the equation of the line $AB$ is

To calculate the equation of the straight line, we first calculate the gradient $m$ of the line:

Secondly, we calculate the value of the $y$-intercept ($c$). We do this by substituing the coordinates of point $A$ into the general form for a straight line:

Therefore, the equation of the line $AB$ is

To calculate the equation of the straight line, we first calculate the gradient $m$ of the line:

Secondly, we calculate the value of the $y$-intercept ($c$). We do this by substituing the coordinates of point $A$ into the general form for a straight line:

Therefore, the equation of the line $AB$ is

To calculate the equation of $AB$, we first need to determine the gradient of the line. We know that if two lines are parallel, then they must have equal gradients:

Next we calculate the value of the $c$ by substituting the coordinates of point $A$ into the equation:

Therefore, the equation of $AB$ is

To calculate the equation of $AB$, we first need to determine the gradient of the line. We know that if two lines are parallel, then they must have equal gradients:

Next we calculate the value of the $c$ by substituting the coordinates of point $A$ into the equation:

Therefore, the equation of $AB$ is

To calculate the equation of $AB$, we first need to determine the gradient of the line. We know that if two lines are parallel, then they must have equal gradients:

Next we calculate the value of the $c$ by substituting the coordinates of point $A$ into the equation:

Therefore, the equation of $AB$ is

Calculate the gradient $m$ using the formula:

where $(x_A;y_A)$ are the coordinates of point $A$ and $(x_B;y_B)$ are the coordinates of the point $B$.

The question asks us to find the gradient of the line shown. We can use the gradient formula to do this.

The gradient formula uses the coordinates of two points (remember that the symbol for gradient is $m$):

Now we substitute the coordinates of the points into the formula to calculate the gradient. Remember to use brackets around the numbers when you substitute to keep all of the negative signs organised.

Therefore the gradient $m$ of the line $AB$ is $\text{1.5}$.

Calculate the gradient $m$ using the formula:

where $(x_A;y_A)$ are the coordinates of point $A$ and $(x_B;y_B)$ are the coordinates of the point $B$.

The question asks us to find the gradient of the line shown. We can use the gradient formula to do this.

The gradient formula uses the coordinates of two points (remember that the symbol for gradient is $m$):

Now we substitute the coordinates of the points into the formula to calculate the gradient. Remember to use brackets around the numbers when you substitute to keep all of the negative signs organised.

Therefore the gradient $m$ of the line $AB$ is $\text{1}$.

Calculate the gradient $m$ using the formula:

where $(x_A;y_A)$ are the coordinates of point $A$ and $(x_B;y_B)$ are the coordinates of the point $B$.

The question asks us to find the gradient of the line shown. We can use the gradient formula to do this.

The gradient formula uses the coordinates of two points (remember that the symbol for gradient is $m$):

Now we substitute the coordinates of the points into the formula to calculate the gradient. Remember to use brackets around the numbers when you substitute to keep all of the negative signs organised.

Therefore the gradient $m$ of the line $AB$ is $\text{1}$.

We have a line passing through two points, but Point $B$ has an unknown coordinate. We can use the equation of the line to find the value of the missing coordinate.

Remember that every point on the line solves the equation (the coordinates of the point make the equation true). So substitute $B \left(x; - \frac{5}{2} \right)$ - the coordinates of $B$ - into the equation and solve for the unknown value. To make it a bit more clear, we can use $x_B$ to show that the coordinate we want belongs to Point $B$.

The $x$-coordinate of Point $B$ is $-3$.

We have a line passing through two points, but Point $B$ has an unknown coordinate. We can use the equation of the line to find the value of the missing coordinate.

Remember that every point on the line solves the equation (the coordinates of the point make the equation true). So substitute $B \left(x; - \frac{1}{2} \right)$ - the coordinates of $B$ - into the equation and solve for the unknown value. To make it a bit more clear, we can use $x_B$ to show that the coordinate we want belongs to Point $B$.

The $x$-coordinate of Point $B$ is $-3$.

We have a line passing through two points, but Point $B$ has an unknown coordinate. We can use the equation of the line to find the value of the missing coordinate.

Remember that every point on the line solves the equation (the coordinates of the point make the equation true). So substitute $B \left(2; y \right)$ - the coordinates of $B$ - into the equation and solve for the unknown value. To make it a bit more clear, we can use $y_B$ to show that the coordinate we want belongs to Point $B$.

The $y$-coordinate of Point $B$ is $- \frac{11}{2}$.

Use the fact that parallel lines have equal gradients.

We need to find the $x$-coordinate of Point $B$. We do not know much about line $AB$. But we do know that it is parallel to the line $y = -2x-5$. That means the lines have the same gradient! We can read the gradient from the equation.

Line $AB$ has the same gradient, so we can start building the equation for the line:

We know the coordinates of Point $A$. We can use them with this equation to calculate the value of $c$.

Great! Now we have the equation for line $AB: y_{AB} = -2x-2$.

Finally, we substitute the coordinates of Point $B$ into the equation for line $AB$:

Solving, we get:

The $x$-coordinate of Point $B$ is $\text{1}$.

Use the fact that parallel lines have equal gradients.

We need to find the $x$-coordinate of Point $B$. We do not know much about line $AB$. But we do know that it is parallel to the line $y = -x-3$. That means the lines have the same gradient! We can read the gradient from the equation.

Line $AB$ has the same gradient, so we can start building the equation for the line:

We know the coordinates of Point $A$. We can use them with this equation to calculate the value of $c$.

Great! Now we have the equation for line $AB: y_{AB} = -x+2$.

Finally, we substitute the coordinates of Point $B$ into the equation for line $AB$:

Solving, we get:

The $x$-coordinate of Point $B$ is $\text{1}$.

Use the fact that parallel lines have equal gradients.

We need to find the $x$-coordinate of Point $B$. We do not know much about line $AB$. But we do know that it is parallel to the line $y = -2x- \frac{7}{2}$. That means the lines have the same gradient! We can read the gradient from the equation.

Line $AB$ has the same gradient, so we can start building the equation for the line:

We know the coordinates of Point $A$. We can use them with this equation to calculate the value of $c$.

Great! Now we have the equation for line $AB: y_{AB} = -2x+\frac{1}{2}$.

Finally, we substitute the coordinates of Point $B$ into the equation for line $AB$:

Solving, we get:

The $x$-coordinate of Point $B$ is $\text{1}$.

Start by using the fact that the two lines must have equal gradients because they are parallel. So you can find the gradient of line $AB$ by looking at the equation for the other line.

The two lines in the graph are parallel. That means they must have equal gradients. The equation for the red dashed line is:

which means that both of the lines have a gradient of $\dfrac{3}{2}$. So the equation of line $AB$ must be:

In this question we need a missing coordinate from point $B$. However, we know both coordinates for point $A$, so we can use them to find the value of $c$ in the equation for line $AB$.

Now we have the complete equation for line $AB$:

Now we can substitute the $x$-coordinate for point $B$ into the equation for line $AB$ and solve for $y_{B}$. This works because the equation of any line agrees with all of the points on the line: if we substitute in the $x$-coordinate then when we solve for $y$ we will get the coordinate for the point which belongs on the line: the equation forces this to be true!

The correct answer is $y_B = \dfrac{7}{2}$.

Start by using the fact that the two lines must have equal gradients because they are parallel. So you can find the gradient of line $AB$ by looking at the equation for the other line.

The two lines in the graph are parallel. That means they must have equal gradients. The equation for the red dashed line is:

which means that both of the lines have a gradient of $\dfrac{1}{2}$. So the equation of line $AB$ must be:

In this question we need a missing coordinate from point $B$. However, we know both coordinates for point $A$, so we can use them to find the value of $c$ in the equation for line $AB$.

Now we have the complete equation for line $AB$:

Now we can substitute the $x$-coordinate for point $B$ into the equation for line $AB$ and solve for $y_{B}$. This works because the equation of any line agrees with all of the points on the line: if we substitute in the $x$-coordinate then when we solve for $y$ we will get the coordinate for the point which belongs on the line: the equation forces this to be true!

The correct answer is $y_B = \dfrac{13}{2}$.

Start by using the fact that the two lines must have equal gradients because they are parallel. So you can find the gradient of line $AB$ by looking at the equation for the other line.

The two lines in the graph are parallel. That means they must have equal gradients. The equation for the red dashed line is:

which means that both of the lines have a gradient of $\dfrac{2}{3}$. So the equation of line $AB$ must be:

In this question we need a missing coordinate from point $A$. However, we know both coordinates for point $B$, so we can use them to find the value of $c$ in the equation for line $AB$.

Now we have the complete equation for line $AB$:

Now we can substitute the $y$-coordinate for point $A$ into the equation for line $AB$ and solve for $x_{A}$. This works because the equation of any line agrees with all of the points on the line: if we substitute in the $y$-coordinate then when we solve for $x$ we will get the coordinate for the point which belongs on the line: the equation forces this to be true!

The correct answer is $x_A = -5$.

You should start with the standard form for a straight line ($y = mx + c$) and substitute in the gradient. Then use the coordinates of point $A$ to find $c$.

The first thing to remember is that a straight line has this general form:

where $m$ is the gradient and $c$ is the $y$-intercept. We have the gradient $m$ of the line (it is given in the question) so we can substitute that into the equation of the line:

Now we need the value of $c$ for the line. We can find this by substituing the coordinates of any point into the general form for a straight line. This works because every point on the line solves the equation (the coordinates of the points agree with the equation). Since we have the coordinates of point $A$ we will use them.

Notice that $c$ is negative: this agrees with the diagram, because the $y$-intercept is below the $x$-axis.

The equation of the line is:

You should start with the standard form for a straight line ($y = mx + c$) and substitute in the gradient. Then use the coordinates of point $A$ to find $c$.

The first thing to remember is that a straight line has this general form:

where $m$ is the gradient and $c$ is the $y$-intercept. We have the gradient $m$ of the line (it is given in the question) so we can substitute that into the equation of the line:

Now we need the value of $c$ for the line. We can find this by substituing the coordinates of any point into the general form for a straight line. This works because every point on the line solves the equation (the coordinates of the points agree with the equation). Since we have the coordinates of point $A$ we will use them.

Notice that $c$ is negative: this agrees with the diagram, because the $y$-intercept is below the $x$-axis.

The equation of the line is:

You should start with the standard form for a straight line ($y = mx + c$) and substitute in the gradient. Then use the coordinates of point $A$ to find $c$.

The first thing to remember is that a straight line has this general form:

where $m$ is the gradient and $c$ is the $y$-intercept. We have the gradient $m$ of the line (it is given in the question) so we can substitute that into the equation of the line:

Now we need the value of $c$ for the line. We can find this by substituing the coordinates of any point into the general form for a straight line. This works because every point on the line solves the equation (the coordinates of the points agree with the equation). Since we have the coordinates of point $A$ we will use them.

Notice that $c$ is negative: this agrees with the diagram, because the $y$-intercept is below the $x$-axis.

The equation of the line is:

Start by calculating the equation of the straight line $AB$.

To calculate the unknown variable on the line, we start by calculating the equation of the straight line. We first calculate the gradient ($m$) of the line $AB$:

Secondly, we calculate the value of the $y$-intercept ($c$) of the line $AB$. We do this by substituting Point $A$ into the general form for a straight line:

We now substitute the known values of Point $C$ into the known equation of the straight line:

Start by calculating the equation of the straight line $AB$.

To calculate the unknown variable on the line, we start by calculating the equation of the straight line. We first calculate the gradient ($m$) of the line $AB$:

Secondly, we calculate the value of the $y$-intercept ($c$) of the line $AB$. We do this by substituting Point $A$ into the general form for a straight line:

We now substitute the known values of Point $C$ into the known equation of the straight line:

Start by calculating the equation of the straight line $AB$.

To calculate the unknown variable on the line, we start by calculating the equation of the straight line. We first calculate the gradient ($m$) of the line $AB$:

Secondly, we calculate the value of the $y$-intercept ($c$) of the line $AB$. We do this by substituting Point $A$ into the general form for a straight line:

We now substitute the known values of Point $C$ into the known equation of the straight line:

Use the equation together with the coordinates of point $B$.

We have two points on a line, but one of the coordinates is missing. We need to find that coordinate. The equation given in the question is the key to finding the answer.

Every point on the line must agree with this equation. That includes the coordinates of point $B$. We substitute the known value for point $B$ into the equation because that will give us the value of the missing coordinate $y$ which agrees with the equation.

Solving, we get:

Use the equation together with the coordinates of point $B$.

We have two points on a line, but one of the coordinates is missing. We need to find that coordinate. The equation given in the question is the key to finding the answer.

Every point on the line must agree with this equation. That includes the coordinates of point $B$. We substitute the known value for point $B$ into the equation because that will give us the value of the missing coordinate $y$ which agrees with the equation.

Solving, we get:

Use the equation together with the coordinates of point $B$.

We have two points on a line, but one of the coordinates is missing. We need to find that coordinate. The equation given in the question is the key to finding the answer.

Every point on the line must agree with this equation. That includes the coordinates of point $B$. We substitute the known value for point $B$ into the equation because that will give us the value of the missing coordinate $x$ which agrees with the equation.

Solving, we get:

Start by finding the gradient of the line. You can take the gradient of the first line straight from the equation: $y = \frac{3 x}{2} + 1$. The gradient for this line is $m = \frac{3}{2}$.

For a line which is parallel to this one, the gradient of the line we want is also $m = \frac{3}{2}$.

Now use the point $(7 ; -5)$, given together with the gradient, to get the equation of the line. Substitute the $x$ and $y$-values into the equation to get the value of $c$:

Therefore, the equation of the line is $y = \frac{3 x}{2} - \frac{31}{2}$.

The graph below shows the two equations: the blue line is the equation given in the question, and the red line is the answer. The point $A( 7 ; -5)$ is shown with a black dot and you can see that the red line passes through the point. You can also see that the two lines are parallel.

Start by finding the gradient of the line. You can take the gradient of the first line straight from the equation: $y = - 2 x - 1$. The gradient for this line is $m = -2$.

For a line which is parallel to this one, the gradient of the line we want is also $m = -2$.

Now use the point $(4 ; -4)$, given together with the gradient, to get the equation of the line. Substitute the $x$ and $y$-values into the equation to get the value of $c$:

Therefore, the equation of the line is $y = - 2 x + 4$.

The graph below shows the two equations: the blue line is the equation given in the question, and the red line is the answer. The point $P( 4 ; -4)$ is shown with a black dot and you can see that the red line passes through the point. You can also see that the two lines are parallel.

Start by finding the gradient of the line. You can take the gradient of the first line straight from the equation: $y = 2 x - 2$. The gradient for this line is $m = 2$.

For a line which is parallel to this one, the gradient of the line we want is also $m = 2$.

Now use the point $(4 ; 0)$, given together with the gradient, to get the equation of the line. Substitute the $x$ and $y$-values into the equation to get the value of $c$:

Therefore, the equation of the line is $y = 2 x - 8$.

The graph below shows the two equations: the blue line is the equation given in the question, and the red line is the answer. The point $N( 4 ; 0)$ is shown with a black dot and you can see that the red line passes through the point. You can also see that the two lines are parallel.

To find the equation of any straight line, we need the gradient of the line. We have two points on the line, so we can use the gradient formula.

So the gradient of the line is $\frac{3}{4}$.

Now we need to find the $y$-intercept of the line. We do this by substituting the gradient and any point on the line into the general form for a straight line. Here we will use the coordinates of point $A$. (You can use point $B$ if you want: any point on the line will work.)

Therefore, the equation of the line $AB$ is as follows:

To find the equation of any straight line, we need the gradient of the line. We have two points on the line, so we can use the gradient formula.

So the gradient of the line is $-2$.

Now we need to find the $y$-intercept of the line. We do this by substituting the gradient and any point on the line into the general form for a straight line. Here we will use the coordinates of point $A$. (You can use point $B$ if you want: any point on the line will work.)

Therefore, the equation of the line $AB$ is as follows:

To find the equation of any straight line, we need the gradient of the line. We have two points on the line, so we can use the gradient formula.

So the gradient of the line is $2$.

Now we need to find the $y$-intercept of the line. We do this by substituting the gradient and any point on the line into the general form for a straight line. Here we will use the coordinates of point $A$. (You can use point $B$ if you want: any point on the line will work.)

Therefore, the equation of the line $AB$ is as follows:

We are given the equation of the straight line:

To determine the value of $x$, we substitute the $y$-coordinate for point $B$ into the equation:

We rearrange the equation to get the answer:

We are given the equation of the straight line:

To determine the value of $y$, we substitute the $x$-coordinate for point $B$ into the equation:

We rearrange the equation to get the answer:

We are given the equation of the straight line:

To determine the value of $x$, we substitute the $y$-coordinate for point $B$ into the equation:

We rearrange the equation to get the answer:

You can only answer this question by comparing the gradients of the two equations, which means you must arrange the equations to be in standard form. For the first equation this is:

This equation is the same as $y = 0 \cdot x + \frac{4}{3}$. Thus the gradient of this equation is $m = 0$.

Now change the second equation to standard form. This is an unusual case: there is no $y$ in the equation! You need to isolate the $x$.

This equation does not show the gradient because it is not in standard form! You need to remember that an equation like this one (which says $x =$ a number) represents a vertical line; and that means the gradient is undefined.

Compare the two gradients to decide if the lines are parallel or not. This is a special situation: one of the lines has a gradient of zero, while the other gradient is undefined. Remember that a horizontal line has a zero gradient and for a vertical line the gradient is undefined. This means that these two lines are perpendicular to each other. Therefore, the correct response is: not parallel.

The two equations are graphed on the Cartesian plane shown below. You can see that there is one horizontal line and one vertical line. In fact, these lines are perpendicular to each other.

You can only answer this question by comparing the gradients of the two equations, which means you must arrange the equations to be in standard form. For the first equation this is:

This equation is the same as $y = 0 \cdot x - \frac{1}{2}$. Thus the gradient of this equation is $m = 0$.

Now change the second equation to standard form. This is an unusual case: there is no $y$ in the equation! You need to isolate the $x$.

This equation does not show the gradient because it is not in standard form! You need to remember that an equation like this one (which says $x =$ a number) represents a vertical line; and that means the gradient is undefined.

Compare the two gradients to decide if the lines are parallel or not. This is a special situation: one of the lines has a gradient of zero, while the other gradient is undefined. Remember that a horizontal line has a zero gradient and for a vertical line the gradient is undefined. This means that these two lines are perpendicular to each other. Therefore, the correct response is: not parallel.

The two equations are graphed on the Cartesian plane shown below. You can see that there is one horizontal line and one vertical line. In fact, these lines are perpendicular to each other.

You can only answer this question by comparing the gradients of the two equations, which means you must arrange the equations to be in standard form. For the first equation this is:

The gradient of the first equation is $m = \frac{5}{3}$.

Now change the second equation to standard form.

The gradient of the second equation is $m = \frac{5}{3}$.

Compare the two gradients to decide if the lines are parallel or not. The equations have equal gradients, so they must be parallel lines. The correct choice is: parallel.

The two equations are graphed on the Cartesian plane shown below. You can see that the two lines run parallel to one another.

Start by using the equation of a straight line, which includes the gradient. Substitute in values that you know.

We have two endpoints of a line segment, and the gradient of the segment. However, the $x$-coordinate of Point $B$ is missing. We need to find that coordinate value.

We can solve this question using the equation for a straight line:

Firstly, we can calculate the value of the $y$-intercept ($c$) of the line $AB$. We do this by substituting the gradient value and the coordinates of Point $A$ into the general form for a straight line:

This means that the equation for the line segment in the graph is:

Now we substitute the known value for Point $B$ into the equation:

Solving, we get:

Start by using the equation of a straight line, which includes the gradient. Substitute in values that you know.

We have two endpoints of a line segment, and the gradient of the segment. However, the $x$-coordinate of Point $B$ is missing. We need to find that coordinate value.

We can solve this question using the equation for a straight line:

Firstly, we can calculate the value of the $y$-intercept ($c$) of the line $AB$. We do this by substituting the gradient value and the coordinates of Point $A$ into the general form for a straight line:

This means that the equation for the line segment in the graph is:

Now we substitute the known value for Point $B$ into the equation:

Solving, we get:

Start by using the equation of a straight line, which includes the gradient. Substitute in values that you know.

We have two endpoints of a line segment, and the gradient of the segment. However, the $y$-coordinate of Point $B$ is missing. We need to find that coordinate value.

We can solve this question using the equation for a straight line:

Firstly, we can calculate the value of the $y$-intercept ($c$) of the line $AB$. We do this by substituting the gradient value and the coordinates of Point $A$ into the general form for a straight line:

This means that the equation for the line segment in the graph is:

Now we substitute the known value for Point $B$ into the equation:

Solving, we get:

Start by recalling that the gradient of parallel lines is equal.

To calculate the equation of the straight line, we first need the gradient ($m$) of the line $AB$. We can get the gradient from the parallel line. The parallel line has (by definition) the same gradient as the line $AB$:

Secondly, we calculate the value of the y-intercept ($c$) of the line $AB$. We do this by substituting a point $A$ into the general form for a straight line:

Therefore, the equation of the line $AB$ is as follows:

Start by recalling that the gradient of parallel lines is equal.

To calculate the equation of the straight line, we first need the gradient ($m$) of the line $AB$. We can get the gradient from the parallel line. The parallel line has (by definition) the same gradient as the line $AB$:

Secondly, we calculate the value of the y-intercept ($c$) of the line $AB$. We do this by substituting a point $A$ into the general form for a straight line:

Therefore, the equation of the line $AB$ is as follows:

Start by recalling that the gradient of parallel lines is equal.

To calculate the equation of the straight line, we first need the gradient ($m$) of the line $AB$. We can get the gradient from the parallel line. The parallel line has (by definition) the same gradient as the line $AB$:

Secondly, we calculate the value of the y-intercept ($c$) of the line $AB$. We do this by substituting a point $A$ into the general form for a straight line:

Therefore, the equation of the line $AB$ is as follows:

The standard formula of a straight line is:

where $m$ is the slope of the graph and $c$ is the point where the graph intercepts the $y$-axis. We are given only two points on the straight line. So we should use those points to try and find the slope and point where the graph intercepts the $y$-axis of the equation. Gradient is basically determined by the ratio of vertical change over horizontal change, 'rise-over-run'. We are not given the graph, but we can always use the gradient formula:

In this specific case we can represent each coordinate as follows:

Now substitute these values into the formula to find the gradient.

If we look at the standard form of a straight line equation again, we now have:

The graph 'cuts' the $y$-axis at $y = - \dfrac{277}{52}$.

The only thing left to do is to give the full equation of the straight line. We have solved all the missing parts that make up the formula of a straight line.

Therefore the equation is

The standard formula of a straight line is:

where $m$ is the gradient of the graph and $c$ is the $y$-intercept. We are given only two points on the straight line. So we should use those points to try and find the slope and $y$-intercept of the equation. Gradient is basically determined by the ratio of vertical change over horizontal change, 'rise-over-run'. We are not given the graph, but we can always use the gradient formula:

In this specific case we can represent each coordinate as follows:

Now substitute these values into the formula to find the gradient.

If we look at the standard form of a straight line equation again, we now have:

The graph 'cuts' the $y$-axis at $y = \dfrac{112}{19}$.

The only thing left to do is to give the full equation of the straight line. We have solved all the missing parts that make up the formula of a straight line.

Therefore the equation is

The standard formula of a straight line is:

where $m$ is the slope of the graph and $c$ is the $y$-intercept. We are given only two points on the straight line. So we should use those points to try and find the gradient and $y$-intercept of the equation. Gradient is basically determined by the ratio of vertical change over horizontal change, 'rise-over-run'. We are not given the graph, but we can always use the gradient formula:

In this specific case we can represent each coordinate as follows:

Now substitute these values into the formula to find the gradient.

If we look at the standard form of a straight line equation again, we now have:

The only thing left to do is to give the full equation of the straight line. We have solved all the missing parts that make up the formula of a straight line.

Therefore the equation is

The coordinates given look a bit complex, but focus on what this question is asking about. Start by writing down the formula for what you need to calculate, and substitute the coordinates into it.

This is a distance formula question: we need to calculate the distance between the points $\left( -2 \; ; \; 2 \right)$ and $\left( 4 \sqrt{3} \; ; \; 4 \sqrt{3} \right)$. On a graph, the points and the distance that we want look like this:

To begin, we need to substitute the coordinates into the formula. It can be helpful to list the coordinates from the points according to the values we need in the distance formula.

Now substitute these values into the formula. Use brackets carefully here to keep everything organised. (There will be lots of brackets, so we will change the brackets in the formula into square brackets to distinguish them from the substituted values.)

Now we follow BODMAS, which means we need to simplify inside the square brackets as much as possible. However, do not change the surds into decimals: often expressions like this simplify after a few steps. Let's see what happens when we expand the square brackets.

Notice that there is a double negative in the first square bracket which changes to addition.

That simplified a lot! The key part is the cancellation of the surd terms, $16 \sqrt{3}$ and $- 16 \sqrt{3}$. From this point on there are only integer values to work with inside the root.

Now we can finish the calculation. Remember, we cannot evaluate the square root until all of the calculations inside of it are done, because a root acts like brackets on the expression under it.

In this case, the expression cannot be simplified any more: it is a surd. The question tells us the answer must exact, so we have the answer. (This surd can be simplied to $2 \sqrt{26}$, which is also accepted.)

The distance between the points is: $\sqrt{104}$.

The coordinates given look a bit complex, but focus on what this question is asking about. Start by writing down the formula for what you need to calculate, and substitute the coordinates into it.

This is a distance formula question: we need to calculate the distance between the points $\left( -2 \; ; \; - 2 \sqrt{10} + 1 \right)$ and $\left( - 2 \sqrt{10} - 2 \; ; \; 1 \right)$. On a graph, the points and the distance that we want look like this:

To begin, we need to substitute the coordinates into the formula. It can be helpful to list the coordinates from the points according to the values we need in the distance formula.

Now substitute these values into the formula. Use brackets carefully here to keep everything organised. (There will be lots of brackets, so we will change the brackets in the formula into square brackets to distinguish them from the substituted values.)

Now we follow BODMAS, which means we need to simplify inside the square brackets as much as possible. However, do not change the surds into decimals: often expressions like this simplify after a few steps. Let's see what happens when we simplify the square brackets.

In this case, we have to distribute the negative sign in the second square bracket into the binomial $- 2 \sqrt{10} + 1$. One of the binomial terms is negative, so it becomes positive and the other term becomes negative. Additionally, there is a double negative in the first square bracket which simplifies to addition.

Brilliant: the trinomials in the square brackets each simplified into a single term and squaring the terms cancelled the roots. From this point on there are only integer values to work with inside the root.

Now we can finish the calculation. Remember, we cannot evaluate the square root until all of the calculations inside of it are done, because a root acts like brackets on the expression under it.

In this case, the expression cannot be simplified any more: it is a surd. The question tells us the answer must exact, so we have the answer. (This surd can be simplied to $4 \sqrt{5}$, which is also accepted.)

The distance between the points is: $\sqrt{80}$.

The coordinates given look a bit complex, but focus on what this question is asking about. Start by writing down the formula for what you need to calculate, and substitute the coordinates into it.

This is a distance formula question: we need to calculate the distance between the points $\left( 3 \; ; \; -3 \right)$ and $\left( - 6 \sqrt{2} \; ; \; - 6 \sqrt{2} \right)$. On a graph, the points and the distance that we want look like this:

To begin, we need to substitute the coordinates into the formula. It can be helpful to list the coordinates from the points according to the values we need in the distance formula.

Now substitute these values into the formula. Use brackets carefully here to keep everything organised. (There will be lots of brackets, so we will change the brackets in the formula into square brackets to distinguish them from the substituted values.)

Now we follow BODMAS, which means we need to simplify inside the square brackets as much as possible. However, do not change the surds into decimals: often expressions like this simplify after a few steps. Let's see what happens when we expand the square brackets.

Notice that there is a double negative in the second square bracket which changes to addition.

That simplified a lot! The key part is the cancellation of the surd terms, $36 \sqrt{2}$ and $- 36 \sqrt{2}$. From this point on there are only integer values to work with inside the root.

Now we can finish the calculation. Remember, we cannot evaluate the square root until all of the calculations inside of it are done, because a root acts like brackets on the expression under it.

In this case, the expression cannot be simplified any more: it is a surd. The question tells us the answer must exact, so we have the answer. (This surd can be simplied to $9 \sqrt{2}$, which is also accepted.)

The distance between the points is: $\sqrt{162}$.

Use the distance formula: $AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$, where $(x_A;y_A)$ are the coordinates of point $A$ and $(x_B;y_B)$ are the coordinates of the point $B$.

First we recall the equation for distance:

Substituting into the equation, we get:

The length of the line segment, rounded to two decimal places, is $\text{6.71}$.

Use the distance formula: $AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$, where $(x_A;y_A)$ are the coordinates of point $A$ and $(x_B;y_B)$ are the coordinates of the point $B$.

First we recall the equation for distance:

Substituting into the equation, we get:

The length of the line segment, rounded to two decimal places, is $\text{6.71}$.

Use the distance formula: $AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$, where $(x_A;y_A)$ are the coordinates of point $A$ and $(x_B;y_B)$ are the coordinates of the point $B$.

First we recall the equation for distance:

Substituting into the equation, we get:

The length of the line segment, rounded to two decimal places, is $\text{7.21}$.

Use the distance formula: $AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$, where $(x_A;y_A)$ are the coordinates of point $A$ and $(x_B;y_B)$ are the coordinates of the point $B$.

First we recall the equation for distance:

Substituting into the equation, we get:

Now we re-arrange, and solve for the value of $y$:

We now have a choice between 2 values for $y$. From the diagram we can see that the appropriate value for this question is $-\text{3.5}$.

Use the distance formula: $AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$, where $(x_A;y_A)$ are the coordinates of point $A$ and $(x_B;y_B)$ are the coordinates of the point $B$.

First we recall the equation for distance:

Substituting into the equation, we get:

Now we re-arrange, and solve for the value of $x$:

We now have a choice between 2 values for $x$. From the diagram we can see that the appropriate value for this question is $\text{2}$.

Use the distance formula: $AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$, where $(x_A;y_A)$ are the coordinates of point $A$ and $(x_B;y_B)$ are the coordinates of the point $B$.

First we recall the equation for distance:

Substituting into the equation, we get:

Now we re-arrange, and solve for the value of $x$:

We now have a choice between 2 values for $x$. From the diagram we can see that the appropriate value for this question is $\text{1}$.

Remember that the distance formula is about how far apart the points are. If you move your finger from point $C$ to $D$, or from $D$ to $C$, would the distance be different?

The question is about substituting coordinates into the distance formula. It gives us two coordinate pairs and asks us this: how do we know which of the points $\left( 9 ; 3 \right)$ and $\left( -6 ; -3 \right)$ correspond to the points $\left( x_{1} ; y_{1} \right)$ and $\left( x_{2} ; y_{2} \right)$ in the formula? In other words, how do we know which of the points is the "starting" point?

Let's step back and look at what the question is about. The diagram below shows points $C$ and $D$ on the Cartesian plane. The distance formula calculates the straight-line distance between the points, as shown.

Now getting back to the question at hand: how do we know which of the given points is $\left( x_{1} \; ; \; y_{1} \right)$? Consider substituting the $x$-coordinates for the points, $\text{9}$ and $\text{−6}$, into the term $\left( x_{2} - x_{1} \right)$: if you substitute $\left( \left( -6 \right) - 9 \right)^{2}$ you get $\left( -15 \right)^{2}$, while if you substitute $\left( 9 - \left( -6 \right) \right)^{2}$ you get $\left( 15 \right)^{2}$. As soon as we calculate the squares, we will get the same (positive) answer either way! The same thing is true for the $y$ calculation. So it does not matter which point you chose for $\left( x_{1} \; ; \; y_{1} \right)$ and $\left( x_{2} \; ; \; y_{2} \right)$ because of the squares in the distance formula will always give us positive values. Both of the substitutions here are correct.

Both of these calculations lead to the correct distance, which is $\sqrt{261}$.

The correct answer choice is: It does not matter which point you use for $\left( x_{1} ; y_{1} \right)$.

Look at the coordinates in the work: are they substituted in the correct places?

The question is about substituting coordinates into the distance formula. It gives us two coordinate pairs and shows the substitution into the formula. We must determine if the substitutions are correct or not.

Let's step back and look at what the question is about. The diagram below shows points $C$ and $D$ on the Cartesian plane. The distance formula calculates the straight-line distance between the points, as shown.

Now back to the substitutions in the question: are the numbers in the correct places? In this case, yes they are. In the distance formula, $\left( x_{1} \; ;\; y_{1} \right)$ represents one of the points, and $\left( x_{2} \; ; \; y_{2} \right)$ represents the other point. Adedapo is using point $C$ as $\left( x_{1} \; ; \; y_{1} \right)$ and $D$ as $\left( x_{2} \; ; \; y_{2} \right)$. He could do it the other way around, and that would be correct too. In other words, both of the following are correct:

Both of these calculations lead to the correct distance, which is $\sqrt{185}$.

The correct answer choice is: Yes, the substitutions are correct.

Look at the coordinates in the work: are they substituted in the correct places?

The question is about substituting coordinates into the distance formula. It gives us two coordinate pairs and shows the substitution into the formula. We must determine if the substitutions are correct or not.

Let's step back and look at what the question is about. The diagram below shows points $C$ and $D$ on the Cartesian plane. The distance formula calculates the straight-line distance between the points, as shown.

Now back to the substitutions in the question: are the numbers in the correct places? Shame, but Akinbode's substitutions are not correct because he mixed up the coordinates. In the distance formula, $\left( x_{1} \; ; \; y_{1} \right)$ represents one of the points, and $\left( x_{2} \; ; \; y_{2} \right)$ represents the other point. While it does not matter if you pick point $C$ or $D$ as $\left( x_{1} \; ; \; y_{1} \right)$, the substitutions must be consistent. Akinbode's substitutions are not consistent: the $x$-coordinates and the $y$-coordinates are substituted backwards compared to each other. Either the numbers $-\text{5}$ and $\text{7}$ should be swapped or the numbers $\text{6}$ and $-\text{7}$ should be swapped. The correct substitution can be either of the following:

Both of these calculations lead to the correct distance, which is $\sqrt{313}$.

You might notice that Akinbode's substitutions are a combination of the two correct versions shown above. Because of that, Akinbode would actually be lucky and get the correct answer. However, this is only because the squares in the formula cancel out his error. Even while he may be lucky to get the correct answer, the substitutions are not fully correct.

The correct answer choice is: No, there is an error.

Start with the mid-point formula:

In this formula, $M$ is the mid-point of two other points. The coordinates in the formula, $\left(x_{1} ; y_{1} \right)$ and $\left(x_{2} ; y_{2} \right)$, are the coordinates of those other points.

In this question, $\left(x_{1} ; y_{1} \right)$ and $\left(x_{2} ; y_{2} \right)$ must be the coordinates of points $A$ and $B$. Therefore:

Now substitute these values into the formula:

Working out the calculations, we get the final answers:

We can compare this answer to the diagram above to check if the answer is reasonable: does the position of point $M$ agree with the coordinates $\left( -1 \; ; \dfrac{1}{2} \right)$? Indeed, that looks quite reasonable!

The coordinates of the mid-point are $\left( -1 \; ; \dfrac{1}{2} \right)$.

Start with the mid-point formula:

In this formula, $M$ is the mid-point of two other points. The coordinates in the formula, $\left(x_{1} ; y_{1} \right)$ and $\left(x_{2} ; y_{2} \right)$, are the coordinates of those other points.

In this question, $\left(x_{1} ; y_{1} \right)$ and $\left(x_{2} ; y_{2} \right)$ must be the coordinates of points $A$ and $B$. Therefore:

Now substitute these values into the formula:

Working out the calculations, we get the final answers:

We can compare this answer to the diagram above to check if the answer is reasonable: does the position of point $M$ agree with the coordinates $\left( 0 \; ; -2 \right)$? Indeed, that looks quite reasonable!

The coordinates of the mid-point are $\left( 0 \; ; -2 \right)$.

Start with the mid-point formula:

In this formula, $M$ is the mid-point of two other points. The coordinates in the formula, $\left(x_{1} ; y_{1} \right)$ and $\left(x_{2} ; y_{2} \right)$, are the coordinates of those other points.

In this question, $\left(x_{1} ; y_{1} \right)$ and $\left(x_{2} ; y_{2} \right)$ must be the coordinates of points $A$ and $B$. Therefore:

Now substitute these values into the formula:

Working out the calculations, we get the final answers:

We can compare this answer to the diagram above to check if the answer is reasonable: does the position of point $M$ agree with the coordinates $\left( \dfrac{1}{2} \; ; - \dfrac{1}{2} \right)$? Indeed, that looks quite reasonable!

The coordinates of the mid-point are $\left( \dfrac{1}{2} \; ; - \dfrac{1}{2} \right)$.

We have two points and we need to find their mid-point. For this we must use the mid-point formula:

In this equation, $M$ represents the mid-point, which has the coordinates $\left( x_{M} ; y_{M} \right)$, while the variables $x_{1}$, etc. refer to the coordinates of the end points. In this case, the end points are the points $Q$ and $R$. We must decide which point to use as point $\text{1}$ and which to use as point $\text{2}$. It does not matter which way we do it, so we will just use the order in which the coordinates were given in the question:

This means:

Plugging these values into the mid-point formula, we get:

Now evaluate the expressions to get the answers. Remember that there are two different expressions in the mid-point formula, one for the $x$ coordinates and the other for the $y$ coordinates. It is helpful to separate the calculations, like this:

The resulting coordinates for the mid-point are $\left( 0 \; ; \; \dfrac{3}{8} \right)$. Notice that the $x$ coordinate of the mid-point is zero, which happens because the $x$ coordinates of $Q$ and $R$ are opposite values.

We can do a simple check to see if this is reasonable by making a quick sketch of the points and the answer. You can see that the coordinates we got do in fact sit directly between points $Q$ and $R$.

The mid-point of $Q$ and $R$ is at: $\left( 0 \; ; \; \dfrac{3}{8} \right)$.

We have two points and we need to find their mid-point. For this we must use the mid-point formula:

In this equation, $M$ represents the mid-point, which has the coordinates $\left( x_{M} ; y_{M} \right)$, while the variables $x_{1}$, etc. refer to the coordinates of the end points. In this case, the end points are the points $Q$ and $R$. We must decide which point to use as point $\text{1}$ and which to use as point $\text{2}$. It does not matter which way we do it, so we will just use the order in which the coordinates were given in the question:

This means:

Plugging these values into the mid-point formula, we get:

Now evaluate the expressions to get the answers. Remember that there are two different expressions in the mid-point formula, one for the $x$ coordinates and the other for the $y$ coordinates. It is helpful to separate the calculations, like this:

The resulting coordinates for the mid-point are $\left( - \dfrac{4}{3} \; ; \; 1 \right)$.

We can do a simple check to see if this is reasonable by making a quick sketch of the points and the answer. You can see that the coordinates we got do in fact sit directly between points $Q$ and $R$.

The mid-point of $Q$ and $R$ is at: $\left( - \dfrac{4}{3} \; ; \; 1 \right)$.

We have two points and we need to find their mid-point. For this we must use the mid-point formula:

In this equation, $M$ represents the mid-point, which has the coordinates $\left( x_{M} ; y_{M} \right)$, while the variables $x_{1}$, etc. refer to the coordinates of the end points. In this case, the end points are the points $Q$ and $R$. We must decide which point to use as point $\text{1}$ and which to use as point $\text{2}$. It does not matter which way we do it, so we will just use the order in which the coordinates were given in the question:

This means:

Plugging these values into the mid-point formula, we get:

Now evaluate the expressions to get the answers. Remember that there are two different expressions in the mid-point formula, one for the $x$ coordinates and the other for the $y$ coordinates. It is helpful to separate the calculations, like this:

The resulting coordinates for the mid-point are $\left( \dfrac{13}{6} \; ; \; \dfrac{5}{4} \right)$.

We can do a simple check to see if this is reasonable by making a quick sketch of the points and the answer. You can see that the coordinates we got do in fact sit directly between points $Q$ and $R$.

The mid-point of $Q$ and $R$ is at: $\left( \dfrac{13}{6} \; ; \; \dfrac{5}{4} \right)$.

The question is about a mid-point: write down the mid-point formula and think about what values you can substitute into it.

In this question we have information about three points: we know the coordinates of one end point and the coordinates of the mid-points, but we do not know the coordinates of Point $P$.

We do not know the coordinates of Point $P$, so we use variables to represent them.

The coordinates are given in terms of $a$ and $b$, but do not let the variables distract you! The question is about a mid-point, and that means a point exactly halfway between two other points. In other words, the situation is something like the picture below, and we can solve it through the mid-point formula.

Substitute in the values we have from above into the mid-point formula. The coordinates we have for the mid-point must go in on the left side.

The result above represents two separate equations: one for $x$ and another for $y$. We can write them separately and solve each one. Begin by multiplying the equations by $\text{2}$ to cancel the denominators.

We cannot simplify the expressions any more, so we are done. The resulting answers are the coordinates for the end Point $P$: $\left( 3 a \; ; \; 2 b + 1 \right)$.

The final answers are:

1. The $x$-coordinate of $P$ is $3 a$.
2. The $y$-coordinate of $P$ is $2 b + 1$.

The question is about a mid-point: write down the mid-point formula and think about what values you can substitute into it.

In this question we have information about three points: we know the coordinates of one end point and the coordinates of the mid-points, but we do not know the coordinates of Point $N$.

We do not know the coordinates of Point $N$, so we use variables to represent them.

The coordinates are given in terms of $x$ and $y$, but do not let the variables distract you! The question is about a mid-point, and that means a point exactly halfway between two other points. In other words, the situation is something like the picture below, and we can solve it through the mid-point formula.

Substitute in the values we have from above into the mid-point formula. The coordinates we have for the mid-point must go in on the left side.

The result above represents two separate equations: one for $x$ and another for $y$. We can write them separately and solve each one. Begin by multiplying the equations by $\text{2}$ to cancel the denominators.

We cannot simplify the expressions any more, so we are done. The resulting answers are the coordinates for the end Point $N$: $\left( 3 x + 4 \; ; \; 3 y + 3 \right)$.

The final answers are:

1. The $x$-coordinate of $N$ is $3 x + 4$.
2. The $y$-coordinate of $N$ is $3 y + 3$.

The question is about a mid-point: write down the mid-point formula and think about what values you can substitute into it.

In this question we have information about three points: we know the coordinates of one end point and the coordinates of the mid-points, but we do not know the coordinates of Point $P$.

We do not know the coordinates of Point $P$, so we use variables to represent them.

The coordinates are given in terms of $x$ and $y$, but do not let the variables distract you! The question is about a mid-point, and that means a point exactly halfway between two other points. In other words, the situation is something like the picture below, and we can solve it through the mid-point formula.

Substitute in the values we have from above into the mid-point formula. The coordinates we have for the mid-point must go in on the left side.

The result above represents two separate equations: one for $x$ and another for $y$. We can write them separately and solve each one. Begin by multiplying the equations by $\text{2}$ to cancel the denominators.

We cannot simplify the expressions any more, so we are done. The resulting answers are the coordinates for the end Point $P$: $\left( 2 x^{2} - 1 \; ; \; - y^{2} \right)$.

The final answers are:

1. The $x$-coordinate of $P$ is $2 x^{2} - 1$.
2. The $y$-coordinate of $P$ is $- y^{2}$.

Since the question includes a mid-point, think about how you can use the mid-point formula.

This question is about a mid-point, which as always is a point exactly between two other points. In this case, however, we only know the coordinates of one of those end points:

We do not know the coordinates of Point $B$, so we use variables to represent them.

Since this question is about a mid-point, and we know the coordinates of some of the points involved, we should approach the question through the mid-point formula:

In this case, we know that the mid-point coordinates are $\left( \dfrac{1}{4} \; ; \; \dfrac{1}{2} \right)$, but we do not know the coordinates of Point $B$. So the equation becomes:

The equation above actually represents two separate equations: one for $x$ and another for $y$. Let's write those two equations separately, and solve! Start by cancelling the denominators in the equations. We do this by multiplying both sides of the equation by the LCD of the fractions in the equation. The $x$ equation has an LCD of $\text{4}$ while the $y$ equation will be multiplied by $\text{2}$. (Notice that when the $\text{4}$ cancels the denominator of $\text{2}$, it leaves a factor of 2 which we must distribute into the binomial.)

We get the answers $x_{B}$ = $2$ and $y_{B}$ = $7$. Compare these numbers to the graph in the question: we can see that the answer is consistent with the position of Point $B$ on the graph.

Point $B$ must be at $\left( 2 \; ; \; 7 \right)$, which means the final answers are:

1. The $x$-coordinate of $B$ is $\text{2}$.
2. The $y$-coordinate of $B$ is $\text{7}$.

Since the question includes a mid-point, think about how you can use the mid-point formula.

This question is about a mid-point, which as always is a point exactly between two other points. In this case, however, we only know the coordinates of one of those end points:

We do not know the coordinates of Point $B$, so we use variables to represent them.

Since this question is about a mid-point, and we know the coordinates of some of the points involved, we should approach the question through the mid-point formula:

In this case, we know that the mid-point coordinates are $\left( - \dfrac{7}{4} \; ; \; 6 \right)$, but we do not know the coordinates of Point $B$. So the equation becomes:

The equation above actually represents two separate equations: one for $x$ and another for $y$. Let's write those two equations separately, and solve! Start by cancelling the denominators in the equations. We do this by multiplying both sides of the equation by the LCD of the fractions in the equation. The $x$ equation has an LCD of $\text{4}$ while the $y$ equation will be multiplied by $\text{2}$. (Notice that when the $\text{4}$ cancels the denominator of $\text{2}$, it leaves a factor of 2 which we must distribute into the binomial.)

We get the answers $x_{B}$ = $2$ and $y_{B}$ = $7$. Compare these numbers to the graph in the question: we can see that the answer is consistent with the position of Point $B$ on the graph.

Point $B$ must be at $\left( 2 \; ; \; 7 \right)$, which means the final answers are:

1. The $x$-coordinate of $B$ is $\text{2}$.
2. The $y$-coordinate of $B$ is $\text{7}$.

Since the question includes a mid-point, think about how you can use the mid-point formula.

This question is about a mid-point, which as always is a point exactly between two other points. In this case, however, we only know the coordinates of one of those end points:

We do not know the coordinates of Point $A$, so we use variables to represent them.

Since this question is about a mid-point, and we know the coordinates of some of the points involved, we should approach the question through the mid-point formula:

In this case, we know that the mid-point coordinates are $\left( \dfrac{1}{4} \; ; \; \dfrac{5}{2} \right)$, but we do not know the coordinates of Point $A$. So the equation becomes:

The equation above actually represents two separate equations: one for $x$ and another for $y$. Let's write those two equations separately, and solve! Start by cancelling the denominators in the equations. We do this by multiplying both sides of the equation by the LCD of the fractions in the equation. The $x$ equation has an LCD of $\text{4}$ while the $y$ equation will be multiplied by $\text{2}$. (Notice that when the $\text{4}$ cancels the denominator of $\text{2}$, it leaves a factor of 2 which we must distribute into the binomial.)

We get the answers $x_{A}$ = $2$ and $y_{A}$ = $7$. Compare these numbers to the graph in the question: we can see that the answer is consistent with the position of Point $A$ on the graph.

Point $A$ must be at $\left( 2 \; ; \; 7 \right)$, which means the final answers are:

1. The $x$-coordinate of $A$ is $\text{2}$.
2. The $y$-coordinate of $A$ is $\text{7}$.

This question is about finding a mid-point. It is complex because the coordinates contain variables, but in the end it is still a mid-point calculation. Start by writing down the information from the question. (Notice that the information "$a, \; b \in \mathbb{N}$" does not matter: for the mid-point it is the coordinate values which we need.)

Remember that this is about a mid-point, which means the situation looks somehow like the picture below. We have no idea where the points actually are, but the situation is something like the picture.

We need to substitute the coordinates above into the mid-point formula because we want to find the mid-point of points $A$ and $B$. The mid-point formula is:

where $x_{1}, y_{1}$, etc. represent the coordinates of the end points ($A$ and $B$). So:

Substitute these expressions into the formula. Make sure you use brackets around the substitutions to help stay organised.

Now we need to simplify as much as possible. Remember that there are two different expressions to evaluate in the mid-point formula, one for $x$ and one for $y$. We will separate these calculations to simplify things.

There we have the answer! The mid-point of the two points is: $\left( 2 a \; ; \; 0 \right)$.

This question is about finding a mid-point. It is complex because the coordinates contain variables, but in the end it is still a mid-point calculation. Start by writing down the information from the question. (Notice that the information "$a, \; b \in \mathbb{N}$" does not matter: for the mid-point it is the coordinate values which we need.)

Remember that this is about a mid-point, which means the situation looks somehow like the picture below. We have no idea where the points actually are, but the situation is something like the picture.

We need to substitute the coordinates above into the mid-point formula because we want to find the mid-point of points $Q$ and $R$. The mid-point formula is:

where $x_{1}, y_{1}$, etc. represent the coordinates of the end points ($Q$ and $R$). So:

Substitute these expressions into the formula. Make sure you use brackets around the substitutions to help stay organised.

Now we need to simplify as much as possible. Remember that there are two different expressions to evaluate in the mid-point formula, one for $x$ and one for $y$. We will separate these calculations to simplify things.

There we have the answer! The mid-point of the two points is: $\left( 3 a - 4 \; ; \; - \dfrac{9}{2} \right)$.

This question is about finding a mid-point. It is complex because the coordinates contain variables, but in the end it is still a mid-point calculation. Start by writing down the information from the question. (Notice that the information "$p, \; q \in \mathbb{Z}$" does not matter: for the mid-point it is the coordinate values which we need.)

Remember that this is about a mid-point, which means the situation looks somehow like the picture below. We have no idea where the points actually are, but the situation is something like the picture.

We need to substitute the coordinates above into the mid-point formula because we want to find the mid-point of points $P$ and $N$. The mid-point formula is:

where $x_{1}, y_{1}$, etc. represent the coordinates of the end points ($P$ and $N$). So:

Substitute these expressions into the formula. Make sure you use brackets around the substitutions to help stay organised.

Now we need to simplify as much as possible. Remember that there are two different expressions to evaluate in the mid-point formula, one for $x$ and one for $y$. We will separate these calculations to simplify things.

There we have the answer! The mid-point of the two points is: $\left( - 3 p \; ; \; - 3 q + \dfrac{3}{2} \right)$.

To find the equation of any straight line, you need a point on the line (which is given in the question) and the gradient of the line! Start by finding the gradient of the line you want.

Start by finding the gradient of the line we want. From the given equation we can read off the gradient: $m = \frac{4}{5}$.

For a line which is perpendicular to this, the gradient must be the negative reciprocal: $m = - \frac{5}{4}$.

Now substitute the coordinates $(-4 ; -6)$, together with the gradient, into the equation to find the value of the $y$-intercept, $c$.

Now we just put all of the pieces together into standard form. The equation that we want is

The two equations are graphed on the Cartesian plane shown below: the blue line is the equation given in the question, and the red line is the answer. The point $( -4 ; -6)$ is shown with a black dot. We can see that the red line passes through the point just like it is supposed to! It is also clear that the lines are perpendicular to each other.

The equation of the line is $y = - \dfrac{5 x}{4} - 11$.

To find the equation of any straight line, you need a point on the line (which is given in the question) and the gradient of the line! Start by finding the gradient of the line you want.

Start by finding the gradient of the line we want. This is a special situation, because the equation given represents a vertical line (equations like '$x =$ a number' are always vertical lines). The gradient of a vertical line is undefined.

For a line which is perpendicular to this, we will get a horizontal line, which must have a gradient of zero.

Now substitute the coordinates $(-1 ; -6)$, together with the gradient, into the equation to find the value of the $y$-intercept, $c$.

Now we just put all of the pieces together into standard form. The equation that we want is

The two equations are graphed on the Cartesian plane shown below: the blue line is the equation given in the question, and the red line is the answer. The point $( -1 ; -6)$ is shown with a black dot. We can see that the red line passes through the point just like it is supposed to! It is also clear that while the original equation (in blue) is vertical, the perpendicular line is horizontal.

The equation of the line is $y = -6$.

To find the equation of any straight line, you need a point on the line (which is given in the question) and the gradient of the line! Start by finding the gradient of the line you want.

Start by finding the gradient of the line we want. From the given equation we can read off the gradient: $m = -2$.

For a line which is parallel to this, we must use the same gradient: the gradient of the line we want is also $m = -2$.

Now substitute the coordinates $(-2 ; 2)$, together with the gradient, into the equation to find the value of the $y$-intercept, $c$.

Now we just put all of the pieces together into standard form. The equation that we want is

The two equations are graphed on the Cartesian plane shown below: the blue line is the equation given in the question, and the red line is the answer. The point $( -2 ; 2)$ is shown with a black dot. We can see that the red line passes through the point just like it is supposed to! It is also clear that the two lines are parallel.

The equation of the line is $y = - 2 x - 2$.

The gradient of the line $AB$ is equal to: $\dfrac{-1}{\text{gradient}_{PL}}$, where PL stands for Perpendicular Line.

To calculate the equation of the straight line, we first need the gradient ($m$) of $\overline{AB}$. We can get the gradient from the blue dashed line.

The gradients of perpendicular lines are negative reciprocals of each other. Therefore the gradient of $\overline{AB}$ must be equal to: $\dfrac{-1}{\textrm{gradient}_{\text{PL}}}$, where PL stands for Perpendicular Line. From the equation given for the blue dashed line, we know that its gradient is $\text{0.5}$.

Secondly, we calculate the value of the y-intercept ($c$) of the line $AB$. We do this by substituting a point $A$ into the general form for a straight line:

Therefore, the equation of the line $AB$ is as follows:

The gradient of the line $AB$ is equal to: $\dfrac{-1}{\text{gradient}_{PL}}$, where PL stands for Perpendicular Line.

To calculate the equation of the straight line, we first need the gradient ($m$) of $\overline{AB}$. We can get the gradient from the blue dashed line.

The gradients of perpendicular lines are negative reciprocals of each other. Therefore the gradient of $\overline{AB}$ must be equal to: $\dfrac{-1}{\textrm{gradient}_{\text{PL}}}$, where PL stands for Perpendicular Line. From the equation given for the blue dashed line, we know that its gradient is $\text{0.5}$.

Secondly, we calculate the value of the y-intercept ($c$) of the line $AB$. We do this by substituting a point $A$ into the general form for a straight line:

Therefore, the equation of the line $AB$ is as follows:

The gradient of the line $AB$ is equal to: $\dfrac{-1}{\text{gradient}_{PL}}$, where PL stands for Perpendicular Line.

To calculate the equation of the straight line, we first need the gradient ($m$) of $\overline{AB}$. We can get the gradient from the blue dashed line.

The gradients of perpendicular lines are negative reciprocals of each other. Therefore the gradient of $\overline{AB}$ must be equal to: $\dfrac{-1}{\textrm{gradient}_{\text{PL}}}$, where PL stands for Perpendicular Line. From the equation given for the blue dashed line, we know that its gradient is $\text{1}$.

Secondly, we calculate the value of the y-intercept ($c$) of the line $AB$. We do this by substituting a point $A$ into the general form for a straight line:

Therefore, the equation of the line $AB$ is as follows:

To calculate the inclination of a straight line, we first need to calculate the gradient:

Now that we know the gradient, we use the following equation to determine the value of $\theta$:

Very important: always check that your calculator is in DEG mode when using the trigonometric function buttons.

If the gradient of the line is negative, we add $180°$ to the above angle to get an obtuse angle. In this case, the gradient is positive.

To calculate the inclination of a straight line, we first need to calculate the gradient:

Now that we know the gradient, we use the following equation to determine the value of $\theta$:

Very important: always check that your calculator is in DEG mode when using the trigonometric function buttons.

If the gradient of the line is negative, we add $180°$ to the above angle to get an obtuse angle. In this case, the gradient is negative. (You can see in the diagram that the angle of inclination is greater than 90 degrees.)

To calculate the inclination of a straight line, we first need to calculate the gradient:

Now that we know the gradient, we use the following equation to determine the value of $\theta$:

Very important: always check that your calculator is in DEG mode when using the trigonometric function buttons.

If the gradient of the line is negative, we add $180°$ to the above angle to get an obtuse angle. In this case, the gradient is positive.

To answer this question, you must know the gradients of each equation. To find the gradients, start by arranging the equations in standard form for linear equations:

We have two equations and need to determine if they represent lines which are parallel, perpendicular or neither.

We can only answer this question by comparing the gradients of the two equations. So we need to arrange the equations in standard form. For the first equation this will be:

The gradient of the first equation is $m_{1} = - \frac{1}{3}$.

Now change the second equation into standard form.

The gradient of the second equation is $m_{2} = \frac{5}{3}$.

Now we can compare the two gradients to decide if the lines are parallel, perpendicular, or neither of those. The key facts to remember are that:

• if the gradients are equal the lines are parallel
• if the gradients are negative reciprocals the lines are perpendicular

For the equations in this question, the gradients are not equal, so the lines are not parallel. Also, the gradients are not negative reciprocals of each other, so the lines are not perpendicular.

The two equations are graphed on the Cartesian plane below. We can see that the lines are neither parallel nor perpendicular.

The correct answer is neither.

To answer this question, you must know the gradients of each equation. To find the gradients, start by arranging the equations in standard form for linear equations:

We have two equations and need to determine if they represent lines which are parallel, perpendicular or neither.

We can only answer this question by comparing the gradients of the two equations. So we need to arrange the equations in standard form. For the first equation this will be:

The gradient of the first equation is $m_{1} = \frac{5}{2}$.

Now change the second equation into standard form.

The gradient of the second equation is $m_{2} = \frac{5}{2}$.

Now we can compare the two gradients to decide if the lines are parallel, perpendicular, or neither of those. The key facts to remember are that:

• if the gradients are equal the lines are parallel
• if the gradients are negative reciprocals the lines are perpendicular

The equations have equal gradients, so they must be parallel lines.

The two equations are graphed on the Cartesian plane below. We can see that the two lines run parallel to one another.

The correct answer is parallel.

To answer this question, you must know the gradients of each equation. To find the gradients, start by arranging the equations in standard form for linear equations:

We have two equations and need to determine if they represent lines which are parallel, perpendicular or neither.

We can only answer this question by comparing the gradients of the two equations. So we need to arrange the equations in standard form. For the first equation this will be:

The gradient of the first equation is $m_{1} = \frac{5}{2}$.

Now change the second equation into standard form.

The gradient of the second equation is $m_{2} = - \frac{2}{5}$.

Now we can compare the two gradients to decide if the lines are parallel, perpendicular, or neither of those. The key facts to remember are that:

• if the gradients are equal the lines are parallel
• if the gradients are negative reciprocals the lines are perpendicular

The two gradients are negative reciprocals of each other. So the lines are perpendicular. (Notice also that the gradients have a product of $-\text{1}$: $\left(\frac{5}{2}\right)\left(- \frac{2}{5} \right)$ is equal to $-\text{1}$, which is always true for perpendicular gradients.)

The two equations are graphed on the Cartesian plane below. We can see that the two lines sit at right angles to each other: they are perpendicular.

The correct answer is perpendicular.

The general form of a straight line equation is $y = mx + c$.

To calculate the equation of the straight line, we first need the gradient ($m$) of $AB$, which we can get from the perpendicular line:

For perpendicular lines, we know that $m_{1} \times m_{2} = -1$:

Next we substitute the coordinates of $A$ into the equation to calculate the value of the $y$-intercept ($c$):

Therefore, the equation of $AB$ is:

The general form of a straight line equation is $y = mx + c$.

To calculate the equation of the straight line, we first need the gradient ($m$) of $AB$, which we can get from the perpendicular line:

For perpendicular lines, we know that $m_{1} \times m_{2} = -1$:

Next we substitute the coordinates of $A$ into the equation to calculate the value of the $y$-intercept ($c$):

Therefore, the equation of $AB$ is:

The general form of a straight line equation is $y = mx + c$.

To calculate the equation of the straight line, we first need the gradient ($m$) of $AB$, which we can get from the perpendicular line:

For perpendicular lines, we know that $m_{1} \times m_{2} = -1$:

Next we substitute the coordinates of $A$ into the equation to calculate the value of the $y$-intercept ($c$):

Therefore, the equation of $AB$ is:

We are given the angle of inclination of the line, which we can use to calculate the gradient:

We know the general form of the equation of a straight line is $y = mx + c$.

Substitute the value for $m$ into the equation:

Next we calculate the value of $c$ by substituting the coordinates of point $A(-2;-3)$ into the equation and simplifying:

Therefore, the equation of the line $AB$ is

We are given the angle of inclination of the line, which we can use to calculate the gradient:

We know the general form of the equation of a straight line is $y = mx + c$.

Substitute the value for $m$ into the equation:

Next we calculate the value of $c$ by substituting the coordinates of point $A(-2;-1.5)$ into the equation and simplifying:

Therefore, the equation of the line $AB$ is

We are given the angle of inclination of the line, which we can use to calculate the gradient:

We know the general form of the equation of a straight line is $y = mx + c$.

Substitute the value for $m$ into the equation:

Next we calculate the value of $c$ by substituting the coordinates of point $A(-1;-1.5)$ into the equation and simplifying:

Therefore, the equation of the line $AB$ is

Start by finding the gradient of the line. From the given equation you can see that the gradient is $m = - \frac{2}{3}$.

For lines that are perpendicular, the gradient must be the negative reciprocal, therefore $m = \frac{3}{2}$.

Substitute the coordinates $(2 ; -3)$ and the gradient into the equation to find the value of $c$.

Therefore, the equation of the line is $y = \frac{3 x}{2} - 6$.

The two equations are graphed on the Cartesian plane shown below: the blue line is the equation given in the question, and the red line is the answer. The point $(2 ; -3)$ is shown with a black dot labelled with an $A$. You can see that the red line passes through the point and that the lines are perpendicular to each other.

Start by finding the gradient of the line. From the given equation you can see that the gradient is $m = -3$.

For lines that are parallel, we know that the gradients are equal so $m = -3$.

Substitute the coordinates $(-2 ; -4)$ and the gradient into the equation to find the value of $c$.

Therefore, the equation of the line is $y = - 3 x - 10$.

The two equations are graphed on the Cartesian plane shown below: the blue line is the equation given in the question, and the red line is the answer. The point $(-2 ; -4)$ is shown with a black dot labelled with an $A$. You can see that the red line passes through the point and that the two lines are parallel.

Start by finding the gradient of the line. From the given equation you can see that the gradient is $m = - \frac{1}{3}$.

For lines that are perpendicular, the gradient must be the negative reciprocal, therefore $m = 3$.

Substitute the coordinates $(-4 ; 5)$ and the gradient into the equation to find the value of $c$.

Therefore, the equation of the line is $y = 3 x + 17$.

The two equations are graphed on the Cartesian plane shown below: the blue line is the equation given in the question, and the red line is the answer. The point $(-4 ; 5)$ is shown with a black dot labelled with an $A$. You can see that the red line passes through the point and that the lines are perpendicular to each other.

The gradient of the line $AB$ is equal to:

where $PL$ stands for Perpendicular Line.

We need to find the equation of line $AB$. So we will need the equation of a straight line:

The line $AB$ has a gradient that is equal to:

where $PL$ stands for Perpendicular Line. The gradient of the perpendicular line is $-\text{2}$, so:

Now we know that the equation of line $AB$ is

Now we substitute the known point $A$ into the equation, to find $c$:

And now we have the complete equation for line $AB$:

We can use the equation for line $AB$ to find the missing coordinate. To do this, substitute the known value for point $B$ into the equation for line $AB$:

Solving, we get:

The correct answer is $x = 4$.

The gradient of the line $AB$ is equal to:

where $PL$ stands for Perpendicular Line.

We need to find the equation of line $AB$. So we will need the equation of a straight line:

The line $AB$ has a gradient that is equal to:

where $PL$ stands for Perpendicular Line. The gradient of the perpendicular line is $-\text{2}$, so:

Now we know that the equation of line $AB$ is

Now we substitute the known point $A$ into the equation, to find $c$:

And now we have the complete equation for line $AB$: